# a certain ruby laser emits 2.5 j pulses of light of wavelenght 6940 angstrom .what is the minimum number of cromium ions in the ruby laser??

Arun
25757 Points
5 years ago
There must be at least that many chromium 3+ ions in the crystal.
n 3.489 10 photons 18 = ×
n photons, because 1 joule per pulse divided by joules per photon gives photons per pulse. 1
E
:=
In each pulse of light, there are
E 2.866 10 Joules − 19 E = ×
h c⋅
λ
:=
The energy of a 694 nm photon is
λ 694 10− 9 c 3 10 := ⋅ 8 h 6.63 10 := ⋅
Emmanuel Sherman
28 Points
3 years ago
The formula says E=hc/$\lambda$
So let h=6.63*10e-34
let c=3*10e8
let $\lambda$=6940*10e-9m
E=hc/$\lambda$
= (6.63*10e-34)(3*10e8)/6940*10e-9m
E=2.86*10e-19
Then n=1/E
=1/2.86*10e-19
n=3.496*10e18 will be the minimum number of Cr3+ in the ruby lazer